Problem: Solve for $x$ : $ 3|x - 3| + 5 = -2|x - 3| + 9 $
Solution: Add $ {2|x - 3|} $ to both sides: $ \begin{eqnarray} 3|x - 3| + 5 &=& -2|x - 3| + 9 \\ \\ { + 2|x - 3|} && { + 2|x - 3|} \\ \\ 5|x - 3| + 5 &=& 9 \end{eqnarray} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} 5|x - 3| + 5 &=& 9 \\ \\ { - 5} &=& { - 5} \\ \\ 5|x - 3| &=& 4 \end{eqnarray} $ Divide both sides by ${5}$ $ \dfrac{5|x - 3|} {{5}} = \dfrac{4} {{5}} $ Simplify: $ |x - 3| = \dfrac{4}{5}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{4}{5} $ or $ x - 3 = \dfrac{4}{5} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{4}{5} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{4}{5} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{4}{5} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $5$ $ x = - \dfrac{4}{5} {+ \dfrac{15}{5}} $ $ x = \dfrac{11}{5} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{4}{5} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{4}{5} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{4}{5} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $5$ $ x = \dfrac{4}{5} {+ \dfrac{15}{5}} $ $ x = \dfrac{19}{5} $ Thus, the correct answer is $x = \dfrac{11}{5} $ or $x = \dfrac{19}{5} $.